A circuit consists of a 12.0-V battery connected to three resistors (44ohm, 17ohm and 110ohm) in series. a. Find the current that flows through the battery. I=__mAb. Find the potential difference across the 44ohm resistor. V1=__Vc. Find the potential difference across the 17ohm resistor. V2=__Vd. Find the potential difference across the 110ohm resistor. V3=__V

a) I = 0.0702A = 70.2mA

b) V1 = 3.09V

c) V2 = 1.19V

d) V3 = 7.72V

Explanation:

Given that the circuit consist of three series resistors.

For resistors arranged in series, the total resistance R can be given as:

R = R1 + R2 + R3

R1 = 44 ohms

R2 = 17 ohms

R3 = 110 ohms.

R = 44 + 17 + 110 = 171 ohms

V = 12 V

a) The current of a circuit is given by;

Potential difference V = current * total resistance

V = IR

Making I the subject of formula,

I = V/R

I = 12/171 = 0.0702A

I = 7.02mA

b) the potential difference across any resistors is given by:

V = IR

Since the arrangement is parallel, the same current flows through each of the resistors.

V1 = IR1

V1 = 0.0702 * 44

V1 = 3.09V

c) applying the same rule as b above:

V2 = IR2

V2 = 0.0702 * 17

V2 = 1.19V

d) applying the same rule as b above.

V3 = IR3

V3 = 0.0702 * 110

V3 = 7.72V

I = 70.2mA

V1 = 3.09V

V2 = 1.19V

V3 = 7.72V

Explanation:

Total resistance for a series connection = R1 + R2 + R3 = 44ohm + 17ohm + 110ohm = 171ohm

From ohm's law

Voltage (V) = current (I) * resistance (R)

I = V/R = 12/171 = 0.0702A = 0.0702*1000mA = 70.2mA

V1 = I*R1 = 0.0702*44 = 3.09V

V2 = I*R2 = 0.0702*17 = 1.19V

V3 = I*R3 = 0.0702*110 = 7.72V